7.1 Example: 2D Elasticity Problem

Cantilever beam subjected to an end shear load

• Compute the displacements using 2D linear elasticity
• (Neglect body forces and accelerations:)

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• Note: This solution is similar to that of Advanced Strength and Applied Elasticity By Ansel C. Ugural, Saul K. Fenster

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Boundary Conditions:

Cantilever beam subjected to an end shear load

• at \(y=\pm \frac{h}{2}\)
  • \({\sigma_{yy}}= 0\)
  • \({\sigma_{xy}}= 0\)
  • (The top and bottom surfaces are traction free \(\left\{\vec{T}\right\} = [\sigma] \left\{\vec{n}\right\} = 0\))

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• at \(x=L\)
  • \({\sigma_{xx}}= 0\)
  • \(\int_{-\frac{h}{2}}^{\frac{h}{2}} {\sigma_{xy}}\, b \, dy = -P\)
  • (Integration of the \({\sigma_{xy}}\) must be the applied point force)

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Cantilever beam subjected to an end shear load

• at \(x=0\)
  • \(u = v = {v {}_{,x}} = 0\)
    • (kinematic boundary conditions)
  • \(\int_{-\frac{h}{2}}^{\frac{h}{2}} {\sigma_{xx}}\, b \, dy = 0\)
    • (The total axial force on the left side zero)
  • \(\int_{-\frac{h}{2}}^{\frac{h}{2}} {\sigma_{xx}}\, b \, y \, dy = P \cdot L\)
    • (integration of \({\sigma_{xx}}\) must result in the applied moment)
  • \(\int_{-\frac{h}{2}}^{\frac{h}{2}} {\sigma_{xy}}\, b \, dy = -P\)
    • (Integration of \({\sigma_{xy}}\) must be the applied point force)

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7.1.1 Summary of the 2D Elasticity Problem

\[\begin{equation} {\sigma_{ij}}\Longrightarrow {\varepsilon_{ij}}\Longrightarrow u_i \end{equation}\]

2D Equilibrium Equations

\[\begin{align} {{\sigma_{xx}} {}_{,x}} + {{\sigma_{yx}} {}_{,y}} + b_x =& \, \rho a_x \\ {{\sigma_{xy}} {}_{,x}} + {{\sigma_{yy}} {}_{,y}} + b_y =& \, \rho a_y \\ \end{align}\]

2D Constitutive Equations (Plane Stress)

\[\begin{equation} \left\{ \begin{array}{c} {\varepsilon_{xx}}\\ {\varepsilon_{yy}}\\ {\varepsilon_{xy}}\\ \end{array} \right\} = \left[ \begin{array}{ccc} \frac{1}{E} & \frac{-\nu}{E} & 0 \\ \frac{-\nu}{E} & \frac{1}{E} & 0 \\ 0 & 0 & \frac{1}{2 G} \\ \end{array} \right] \left\{ \begin{array}{c} {\sigma_{xx}}\\ {\sigma_{yy}}\\ {\sigma_{xy}}\\ \end{array} \right\} \end{equation}\]

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2D linearized strain-displacement relationship

\[\begin{align} {\varepsilon_{xx}}=& \, {u {}_{,x}} \\ {\varepsilon_{yy}}=& \, {v {}_{,y}} \\ {\gamma_{xy}}=& \, {u {}_{,y}} + {v {}_{,x}} \\ \end{align}\]

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Solution approach: make reasonable assumptions and manipulate the equations into a tractable solution form.

• Since the beam is not thick, assume \({\sigma_{yy}}= 0\)
  • using equilibrium: \({{\sigma_{xy}} {}_{,x}} + {{\sigma_{yy}} {}_{,y}} = \rho a_y\)
  • thus \({{\sigma_{yy}} {}_{,y}}=0 \Longrightarrow {{\sigma_{xy}} {}_{,x}}=0\)
• Also, by differentiating the equilibrium equations and neglecting body forces and accelerations: \[\begin{equation} \tag{7.1} \begin{split} {{\sigma_{xx}} {}_{,xx}} + {{\sigma_{yx}} {}_{,yx}} =& \, 0 \\ {{\sigma_{yx}} {}_{,xy}} + {{\sigma_{yy}} {}_{,yy}} =& \, 0 \\ \end{split} \end{equation}\]

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• Since the order of differentiation is arbitrary, we solve these for \({{\sigma_{yx}} {}_{,xy}}\)
  • \(\Longrightarrow {{\sigma_{xx}} {}_{,xx}} = {{\sigma_{yy}} {}_{,yy}}\) for plane stress and negligible body forces by substituting equation (7.1)a into (7.1)b and because \({{\sigma_{yy}} {}_{,yy}}=0\)
• Therefore, \({{\sigma_{xx}} {}_{,xx}}=0\)

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\({{\sigma_{xx}} {}_{,xx}}=0\)

• For that result, assume a stress of the form:
  • \({\sigma_{xx}}= \left(C_1 + C_2 \cdot x\right) \cdot f(y) + g(y)\) which is a general form
  • For this case, assume: \({\sigma_{xx}}= \left(C_1 + C_2 \cdot x\right) \cdot y\) because we know the bending moment varies linearly with \(x\) and \({\sigma_{xx}}\) depends linearly on \(y\)
  • To accept it, we must confirm a stress of this form satisfies the boundary conditions

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• Using equilibrium
  • \({{\sigma_{xx}} {}_{,x}} + {{\sigma_{xy}} {}_{,y}} = 0\)
  • \({{\sigma_{xx}} {}_{,x}} = C_2 y\)
  • \(\Longrightarrow {{\sigma_{xy}} {}_{,y}} = -C_2 y\)
• Integrating:
  • \({\sigma_{xy}}= - {\frac{1}{2}}C_2 y^2 + f(x) = - {\frac{1}{2}}C_2 y^2 + C_3\)
  • We know \(f(x)=C_3\), i.e., \(f(x)\) is the same for all cross sections or the 2\(^{\mathrm{nd}}\) equilibrium equation is violated \({{\sigma_{xy}} {}_{,x}}=0\)

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• We can confirm the differential equations (equilibrium) are satisfied with the stress forms we have assumed:
  • \({\sigma_{xx}}= \left(C_1 + C_2 \cdot x\right) \cdot y\)
  • \({\sigma_{yy}}=0\)
  • \({\sigma_{xy}}= -{\frac{1}{2}}C_2 y^2 + C_3\)

Check the boundary conditions:

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• \({\sigma_{xy}}= 0\) at \(y=\pm \frac{h}{2} \Longrightarrow - {\frac{1}{2}}C_2 \left(\frac{h}{2}\right)^2 + C_3 = 0\)
  • Therefore, \(C_3 = C_2 \cdot \frac{h^2}{8}\)
• \({\sigma_{xx}}=0\) at \(x=L \Longrightarrow \left(C_1+C_2 \cdot L\right) \cdot y = 0\)
  • thus, \(C_1 = -C_2 L\)
• Hence:
  • \({\sigma_{xx}}= \left(C_1 + C_2 \cdot x\right) \cdot y = C_2 \cdot \left(x-L\right) \cdot y\)
  • \({\sigma_{xy}}= - {\frac{1}{2}}C_2 y^2 + C^3 = C_2 \cdot \left(\frac{h^2}{8} - \frac{y^2}{2}\right)\)
• At \(x=L\):
  • \(\int_{-\frac{h}{2}}^{\frac{h}{2}} {\sigma_{xy}}\, b \, dy = -P\)
  • \(-P = C_2 \cdot b \cdot \left(\frac{h^3}{8} - \frac{h^3}{24}\right) = C_2 \cdot b \cdot \frac{h^3}{12} = C_2 \cdot I \Longrightarrow C_2 = -\frac{P}{I}\)

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Hence:

• \({\sigma_{xx}}= -\frac{P}{I} \cdot \left(x-L\right) \cdot y\)
• \({\sigma_{yy}}= 0\)
• \({\sigma_{xy}}= -\frac{P}{I} \cdot \left(\frac{h^2}{8} - \frac{y^2}{2}\right)\)

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• To obtain the displacements, we use the stress–strain equations (constitutive) and the strain–displacement (kinematic) equations \[\begin{equation} {\sigma_{yy}}= 0 \end{equation}\]

\[\begin{equation} {\varepsilon_{xx}}=\frac{{\sigma_{xx}}}{E} = - \frac{P}{EI} \cdot \left(x-L\right) \cdot y = {u {}_{,x}} \end{equation}\]

• Therefore: \[\begin{equation} \tag{7.2} u (x,y) = - \frac{P}{EI} \cdot \left(\frac{x^2}{2}-Lx\right) \cdot y + f_1(y) \end{equation}\]

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• also: \[\begin{equation} {\varepsilon_{yy}}=\frac{- \nu {\sigma_{xx}}}{E} = \nu \frac{P}{EI} \cdot \left(x-L\right) \cdot y = {v {}_{,y}} \end{equation}\]

• Therefore: \[\begin{equation} \tag{7.3} v (x,y) = \nu \frac{P}{EI} \cdot \left(x-L\right) \cdot \frac{y^2}{2} + f_2(x) \end{equation}\]

\[\begin{align} 2 {\varepsilon_{xy}}= \frac{{\sigma_{xy}}}{G} =& \, -\frac{P}{GI} \left(\frac{h^2}{8} - y^2\right) \\ =& \, {u {}_{,y}} + {v {}_{,x}} \\ =& \, - \frac{P}{EI} \cdot \left(\frac{x^2}{2}-Lx\right) + \frac{df_1(y)}{dy} + \nu \frac{P}{EI} \cdot \frac{y^2}{2} + \frac{df_2(x)}{dx} \\ \end{align}\]

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• Rearranging the equations: \[\begin{equation} -\frac{P}{GI} \cdot \frac{h^2}{8} = -\frac{P}{EI} \cdot \left(\frac{x^2}{2} - Lx\right) + \frac{df_2(x)}{dx} + P \left(\frac{\nu}{EI} -\frac{1}{GI}\right) \cdot \frac{y^2}{2} + \frac{df_1(y)}{dy} \end{equation}\]

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Rearranging the equations (examine by row):

\[\begin{align} \tag{7.4} \\ -\frac{P}{GI} \cdot \frac{h^2}{8} =& \, \mathrm{constant} \\ = & \, -\frac{P}{EI} \cdot \left(\frac{x^2}{2} - Lx\right) + \frac{df_2(x)}{dx} \\ & \, + P \left(\frac{\nu}{EI} -\frac{1}{GI}\right) \cdot \frac{y^2}{2} + \frac{df_1(y)}{dy} \\ =& \, B_1 + B_2 \\ \end{align}\]

• These are grouped so that parts dependent on \(x\) and parts dependent on \(y\) are isolated as \(B_1\) and \(B_2\)

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• \(B_1\) and \(B_2\) must each also be constant or a change in only \(x\) or only \(y\) or will violate the constancy of the left hand side of (7.4).

  \[\begin{align} B_1 = & \, -\frac{P}{EI} \cdot \left(\frac{x^2}{2} - Lx\right) + \frac{df_2(x)}{dx} \\ B_2 = & \, + P \left(\frac{\nu}{EI}-\frac{1}{GI}\right) \cdot \frac{y^2}{2} + \frac{df_1(y)}{dy} \\ \end{align}\]

Solve for the derivative of the unknown function and integrate: \[\begin{align} \frac{df_2(x)}{dx} =& \, \frac{P}{EI} \cdot \left(\frac{x^2}{2} - Lx\right) + B_1\\ f_2(x) =& \, \frac{P}{EI} \cdot \left(\frac{x^3}{6} - \frac{L x^2}{2}\right) + B_1 x + B_3\\ \end{align}\]

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Recall the equation for \(v\) displacement, i.e., equation (7.3): \[\begin{align} v (x,y) =& \, \nu \frac{P}{EI} \cdot \left(x-L\right) \cdot \frac{y^2}{2} + f_2(x) \\ \end{align}\]

Again, solve for the derivative of the unknown function and integrate: \[\begin{align} \frac{df_1(y)}{dy} =& \, - P \left(\frac{\nu}{EI} -\frac{1}{GI}\right) \cdot \frac{y^2}{2} + B_2 \\ f_1(y) =& \, - P \left(\frac{\nu}{EI} -\frac{1}{GI}\right) \cdot \frac{y^3}{6} + B_2 y + B_4 \\ \end{align}\]

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Recall the equation for \(u\) displacement, i.e., equation (7.2): \[\begin{align} u (x,y) =& \, - \frac{P}{EI} \cdot \left(\frac{x^2}{2}-Lx\right) \cdot y + f_1(y) \\ \end{align}\]

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7.1.2 The displacements are thus:

\[\begin{align} u (x,y) =& \, - \frac{P}{EI} \cdot \left(\frac{x^2}{2}-Lx\right) \cdot y + \frac{P}{EI} \left(2 + \nu\right) \cdot \frac{y^3}{6} + B_2 y + B_4\\ v (x,y) =& \, + \frac{P \nu}{EI} \cdot \left(x-L\right) \cdot \frac{y^2}{2} + \frac{P}{EI} \cdot \left(\frac{x^3}{6} - \frac{L x^2}{2}\right) + B_1 x + B_3 \\ \end{align}\]

Note: \(G = \frac{E}{2(1+\nu)}\) has been used.

There are still 4 unknown constants, must use the boundary conditions to obtain them.

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• Boundary Conditions
  • at \(x=0,y=0\)
    • \(u=0 \Longrightarrow B_4=0\)
    • \(v=0 \Longrightarrow B_3=0\)
    • \({v {}_{,x}}=0 \Longrightarrow B_1=0\)
  • Also, \(B_1+B_2=\, \mathrm{constant}\)
    • \(B_2 = -\frac{P}{GI}\cdot \frac{h^2}{8}\)

Note: at \(x=L,y=0 \Longrightarrow v=-\frac{PL^3}{3 EI}\). This is the same result as we will find after we look at [beam theory] solutions. Once formulated, they will be much easier to apply.

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We have seen how difficult these “simple” solutions in elasticity can be. Engineering formulations such as beam theory are used to obtain approximate solutions that are tractable yet accurate. We will discuss some of these over the remainder of the course.

Lecuture “10” is empty and is a placeholder for an future advanced topic. AE4630 students, please skip to lecture 11.